Additive function

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See Halmos' measure theory book too

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Needs to include everything the old page did, link to propositions and lead to measures

Definition

[math]\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]A real valued set function on a class of sets, [ilmath]\mathcal{A} [/ilmath], [ilmath]f:\mathcal{A}\rightarrow\mathbb{R} [/ilmath] is called additive or finitely additive if:

• For [ilmath]A,B\in\mathcal{A} [/ilmath] with [ilmath]A\cap B=\emptyset[/ilmath] (pairwise disjoint) and [ilmath]A\udot B\in\mathcal{A} [/ilmath] we have:
  • [ilmath]f(A\udot B)=f(A)+f(B)[/ilmath]

Finitely additive

With the same definition of [ilmath]f[/ilmath], we say that [ilmath]f[/ilmath] is finitely additive if for a pairwise disjoint family of sets [ilmath]\{A_i\}_{i=1}^n\subseteq\mathcal{A}[/ilmath] with [ilmath]\bigudot_{i=1}^nA_i\in\mathcal{A}[/ilmath] we have:

• [math]f\left(\mathop{\bigudot}_{i=1}^nA_i\right)=\sum^n_{i=1}f(A_i)[/math].

Claim 1: [ilmath]f[/ilmath] is finitely additive [ilmath]\iff[/ilmath] it is additive

Countably additive

With the same definition of [ilmath]f[/ilmath], we say that [ilmath]f[/ilmath] is countably additive if for a pairwise disjoint family of sets [ilmath]\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}[/ilmath] with [ilmath]\bigudot_{n=1}^\infty A_n\in\mathcal{A}[/ilmath] we have:

• [math]f\left(\mathop{\bigudot}_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}f(A_n)[/math].

Immediate properties

Claim: if [ilmath]\emptyset\in\mathcal{A} [/ilmath] then [ilmath]f(\emptyset)=0[/ilmath]

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Let [ilmath]\emptyset,A\in\mathcal{A} [/ilmath], then:

• [ilmath]f(A)=f(A\udot\emptyset)=f(A)+f(\emptyset)[/ilmath] by hypothesis.
• Thus {{M|1=f(A)=f(A)+f(\emptyset)}]
• This means [ilmath]f(A)-f(A)=f(\emptyset)[/ilmath]

We see [ilmath]f(\emptyset)=0[/ilmath], as required

Proof of claims

Claim 1: [ilmath]f[/ilmath] is additive if and only if [ilmath]f[/ilmath] is finitely additive

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Additive function/Claim 1 - additive iff finitely additive

See also

• Subadditive function

References

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TODO: Check algebra books for definition of additive, perhaps split into two cases, additive set function and additive function

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OLD PAGE

An additive function is a homomorphism that preserves the operation of addition in place on the structure in question.

In group theory (because there's only one operation) it is usually just called a "group homomorphism"

Definition

Here [ilmath](X,+_X:X\times X\rightarrow X)[/ilmath] (which we'll denote [ilmath]X[/ilmath] and [ilmath]+_X[/ilmath]) denotes a set endowed with a binary operation called addition.

The same goes for [ilmath](Y,+_Y:Y\times Y\rightarrow Y)[/ilmath].

A function [ilmath]f[/ilmath] is additive if for [ilmath]a,b\in X[/ilmath]

[math]f(a+_Xb)=f(a)+_Yf(b)[/math]

Warning about structure

If the spaces X and Y have some sort of structure (example: Group) then some required properties follow, for example:

[math]x=x+0\implies f(x)+0=f(x)=f(x+0)=f(x)+f(0)\implies f(0)=0[/math] so one must be careful!

On set functions

A set function, [ilmath]\mu[/ilmath], is called additive if whenever:

• [ilmath]A\in X[/ilmath]
• [ilmath]B\in X[/ilmath]
• [ilmath]A\cap B=\emptyset[/ilmath]

We have:

[math]\mu(A\cup B)=\mu(A)+\mu(B)[/math] for valued set functions (set functions that map to values)

A shorter notation: [math]\mu(A\uplus B)=\mu(A)+\mu(B)[/math], where [ilmath]\uplus[/ilmath] denotes "disjoint union" -- just the union when the sets are disjoint, otherwise undefined.

An example would be a measure.

Variations

Finitely additive

This follows by induction on the additive property above. It states that:

• [math]f\Big(\sum^n_{i=1}A_i\Big)=\sum^n_{i=1}f(A_i)[/math] for additive functions
• [math]\mu\Big(\biguplus^n_{i=1}A_i\Big)=\sum^n_{i=1}\mu(A_i)[/math] for valued set functions

Countably additive

This is a separate property, while given additivity we can get finite additivity, but we cannot get countable additivity from just additivity.

• [math]f\Big(\sum^\infty_{n=1}A_n\Big)=\sum^\infty_{n=1}f(A_n)[/math] for additive functions
• [math]\mu\Big(\biguplus^\infty_{n=1}A_n\Big)=\sum^\infty_{n=1}\mu(A_n)[/math] for valued set functions

Countable additivity can imply additivity

If [math]f(0)=0[/math] or [math]\mu(\emptyset)=0[/math] then given a finite set [math]\{a_i\}_{i=1}^n[/math] we can define an infinite set [math]\{b_n\}_{n=1}^\infty[/math] by:

[math]b_i=\left\{\begin{array}a_i&\text{if }i\le n\\ 0\text{ or }\emptyset & \text{otherwise}\end{array}\right.[/math]

Thus:

• [math]f(\sum^\infty_{n=1}b_n)= \begin{array}{lr} f(\sum^n_{i=1}a_i) \\ \sum^\infty_{n=1}f(b_n)=\sum^n_{i=1}f(a_i)+f(0)=\sum^n_{i=1}f(a_i) \end{array}[/math]
• Or indeed [math]\mu(\sum^\infty_{n=1}b_n)= \begin{array}{lr} \mu(\sum^n_{i=1}a_i) \\ \sum^\infty_{n=1}\mu(b_n)=\sum^n_{i=1}\mu(a_i)+\mu(0)=\sum^n_{i=1}\mu(a_i) \end{array}[/math]